Manipulating URL/array to remove the last element
$begingroup$
Is there a better way to do this using chaining?
var url = "www.mycompany.com/sites/demo/t1"
var x = url.split('/');
console.log(x);
var y = x.pop();
console.log(y,x);
var z = x.join("/");
console.log(z);I tried something like but wouldn't work since pop just returns the last value and not the rest:
parentUrl = self.attr('Url').split("/").pop().join("/");
javascript jquery array url
edited 48 mins ago
200_success
130k16153419
asked Dec 1 '14 at 21:04
BatmanBatman
1034
$endgroup$
add a comment |
$begingroup$
Is there a better way to do this using chaining?
var url = "www.mycompany.com/sites/demo/t1"
var x = url.split('/');
console.log(x);
var y = x.pop();
console.log(y,x);
var z = x.join("/");
console.log(z);I tried something like but wouldn't work since pop just returns the last value and not the rest:
parentUrl = self.attr('Url').split("/").pop().join("/");
javascript jquery array url
edited 48 mins ago
200_success
130k16153419
asked Dec 1 '14 at 21:04
BatmanBatman
1034
$endgroup$
add a comment |
$begingroup$
Is there a better way to do this using chaining?
var url = "www.mycompany.com/sites/demo/t1"
var x = url.split('/');
console.log(x);
var y = x.pop();
console.log(y,x);
var z = x.join("/");
console.log(z);I tried something like but wouldn't work since pop just returns the last value and not the rest:
parentUrl = self.attr('Url').split("/").pop().join("/");
javascript jquery array url
edited 48 mins ago
200_success
130k16153419
asked Dec 1 '14 at 21:04
BatmanBatman
1034
$endgroup$
Is there a better way to do this using chaining?
var url = "www.mycompany.com/sites/demo/t1"
var x = url.split('/');
console.log(x);
var y = x.pop();
console.log(y,x);
var z = x.join("/");
console.log(z);I tried something like but wouldn't work since pop just returns the last value and not the rest:
parentUrl = self.attr('Url').split("/").pop().join("/");
var url = "www.mycompany.com/sites/demo/t1"
var x = url.split('/');
console.log(x);
var y = x.pop();
console.log(y,x);
var z = x.join("/");
console.log(z);var url = "www.mycompany.com/sites/demo/t1"
var x = url.split('/');
console.log(x);
var y = x.pop();
console.log(y,x);
var z = x.join("/");
console.log(z);javascript jquery array url
javascript jquery array url
edited 48 mins ago
200_success
130k16153419
asked Dec 1 '14 at 21:04
BatmanBatman
1034
edited 48 mins ago
200_success
130k16153419
asked Dec 1 '14 at 21:04
BatmanBatman
1034
edited 48 mins ago
200_success
130k16153419
edited 48 mins ago
200_success
130k16153419
edited 48 mins ago
200_success
130k16153419
130k16153419
asked Dec 1 '14 at 21:04
BatmanBatman
1034
asked Dec 1 '14 at 21:04
BatmanBatman
1034
asked Dec 1 '14 at 21:04
BatmanBatman
1034
1034
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
pop() mutates the underlying list: it removes the last item (and returns it).
It seems you're looking for slice:
"www.mycompany.com/sites/demo/t1".split('/').slice(0, -1).join('/')
// -> gives: "www.mycompany.com/sites/demo"
.slice(0, -1) gives the elements of the list from the start until the end minus one item.
However, this is a very poor solution to drop the last path element from a URL. In particular, splitting a string to an array, drop the last element and join the string again is inefficient. It would be better to use lastIndexOf to find the position of the last /, and then substr to extract the part of the string before the last /.
var s = "www.mycompany.com/sites/demo/t1";
s.substr(0, s.lastIndexOf('/'));
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
pop() mutates the underlying list: it removes the last item (and returns it).
It seems you're looking for slice:
"www.mycompany.com/sites/demo/t1".split('/').slice(0, -1).join('/')
// -> gives: "www.mycompany.com/sites/demo"
.slice(0, -1) gives the elements of the list from the start until the end minus one item.
However, this is a very poor solution to drop the last path element from a URL. In particular, splitting a string to an array, drop the last element and join the string again is inefficient. It would be better to use lastIndexOf to find the position of the last /, and then substr to extract the part of the string before the last /.
var s = "www.mycompany.com/sites/demo/t1";
s.substr(0, s.lastIndexOf('/'));
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
$endgroup$
add a comment |
$begingroup$
pop() mutates the underlying list: it removes the last item (and returns it).
It seems you're looking for slice:
"www.mycompany.com/sites/demo/t1".split('/').slice(0, -1).join('/')
// -> gives: "www.mycompany.com/sites/demo"
.slice(0, -1) gives the elements of the list from the start until the end minus one item.
However, this is a very poor solution to drop the last path element from a URL. In particular, splitting a string to an array, drop the last element and join the string again is inefficient. It would be better to use lastIndexOf to find the position of the last /, and then substr to extract the part of the string before the last /.
var s = "www.mycompany.com/sites/demo/t1";
s.substr(0, s.lastIndexOf('/'));
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
$endgroup$
add a comment |
$begingroup$
pop() mutates the underlying list: it removes the last item (and returns it).
It seems you're looking for slice:
"www.mycompany.com/sites/demo/t1".split('/').slice(0, -1).join('/')
// -> gives: "www.mycompany.com/sites/demo"
.slice(0, -1) gives the elements of the list from the start until the end minus one item.
However, this is a very poor solution to drop the last path element from a URL. In particular, splitting a string to an array, drop the last element and join the string again is inefficient. It would be better to use lastIndexOf to find the position of the last /, and then substr to extract the part of the string before the last /.
var s = "www.mycompany.com/sites/demo/t1";
s.substr(0, s.lastIndexOf('/'));
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
$endgroup$
pop() mutates the underlying list: it removes the last item (and returns it).
It seems you're looking for slice:
"www.mycompany.com/sites/demo/t1".split('/').slice(0, -1).join('/')
// -> gives: "www.mycompany.com/sites/demo"
.slice(0, -1) gives the elements of the list from the start until the end minus one item.
However, this is a very poor solution to drop the last path element from a URL. In particular, splitting a string to an array, drop the last element and join the string again is inefficient. It would be better to use lastIndexOf to find the position of the last /, and then substr to extract the part of the string before the last /.
var s = "www.mycompany.com/sites/demo/t1";
s.substr(0, s.lastIndexOf('/'));
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
edited 1 hour ago
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
answered Dec 1 '14 at 21:22
janosjanos
98.4k12125350
98.4k12125350
add a comment |
add a comment |
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